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Unformatted text preview: Solutions to problem set 3 (141A Sp07) 1 1. (ISSP 3.3) Basis of two unalike atoms. At k = π/ a , equation (20) becomes 2 ω 2 M 1 C ! u = 2 ω 2 M 2 C ! v = Note that the u and v sublattices are decoupled at this value of k ; u can change without changing v , and vice versa. There are two solutions to the above equations: u = 0, ω 2 = 2 C / M 2 and v arbitrary; and v = 0, ω 2 = 2 C / M 1 and u arbitrary. 2. (ISSP 3.4) Kohn anomaly. Equation (16a) gives ω 2 = 2 A M ∞ X p = 1 sin pk a pa (1 cos( pka )) = 2 A Ma ∞ X p = 1 sin pk a p 1 2 sin p ( k + k ) a p 1 2 sin p ( k k ) a p ! Gradshteyn and Ryzhik ( Table of integrals, series and products )1.441 gives ∞ X p = 1 sin α p p = π α 2 < α < 2 π (you can prove this by writing the sin as the sum of two exponentials, and then using the taylor expansion of log (1 + x ).) So for k > k , we have ω 2 = 2 A Ma ∞ X p = 1 sin pk a p 1 2 sin p ( k + k ) a p 1 2 sin p ( k k ) a p ! = 2 A Ma π k a 2 1 2 π ( k + k ) a 2 1 2 π ( k k ) a 2 !...
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 Spring '08
 SOUZA
 Derivative, Solid State Physics, Sin, C2 Greyhound, Massachusetts Route 2A, nearest neighbour coupling

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